= Solution
Work over the reals with $x\ne0$, and put $p=dy/dx$. The algebraic equation implies $p\ne0$ and
$$
y=\frac x2\left(p-\frac1p\right).
$$
Differentiate this identity and use $y'=p$; cancellation gives $dp/dx=p/x$. Hence $p=Cx$, with nonzero real $C$, and substitution gives the complete family
$$
\boxed{y=\frac C2x^2-\frac1{2C},\qquad C\ne0.}
$$
There is no extra real singular branch lost in the cancellation, since $p^2+1\ne0$. At a point $(x_0,y_0)$ with $x_0\ne0$, the permitted constants satisfy $x_0^2C^2-2y_0C-1=0$, giving
$$
C_\pm=\frac{y_0\pm\sqrt{x_0^2+y_0^2}}{x_0^2}.
$$
These are distinct, real and nonzero, so \b[exactly two integral curves pass through each such point]. This does not contradict an <Picard-Lindelöf theorem>: the implicit equation permits two different slopes and is not one specified explicit <first-order ordinary differential equation>. Each choice of the smooth slope branch separately defines a locally unique curve.
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